From: "Dr. Sam Lomonaco"
Date: Sun, 17 Oct 1999 00:06:49 -0400 (EDT)
To: ????
Subject: Re: CMSC 442 Homework #4
Cc: lomonaco@umbc.edu

????:

>
>Well, I *thought* I understood everything we'd done in class, but I am
>having a bit of trouble with problem 1, and *lots* of trouble with
>problem 3.
>

u lies in the ideal (x^2 + x + 1) iff there is an element   a  in the ring
R_3 such that  u=a*(x^2 + x + 1).  The elements of R_3 can all be represented
in terms of polynomials of degree < 3, since x^3 = 1.

Therefore, a = (a_2)x^2 + (a_1)x + a_0, where the a_k's are either zero or one.
This using the relation x^3 = 1, you can with a little calculation show that

a*(x^2 + x + 1) = (a_2 + a_1 + a_0)*(x^2 + x + 1)

Thus, (x^2 + x + 1) = { 0, x^2 + x + 1 }

>
>I'm not quite sure how to compute the orders for ksi^i i = 0,1,....,62.
>

The order of  a  is the smallest positive integer  k  such that  a^k = 1
An element  ksi^i  is primitive if it has maximal order  (2^6)-1.  In other
words, if the powers of (ksi^i)^k, k=0..(2^6)-2 are all the non-zero elements
of GF(2^6).

.....

Regards, Dr. L
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Samuel J. Lomonaco   | Univ. of Maryland Baltimore County (UMBC) |
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Email:               | 410-455-3500 (Sec)   |  410-455-2338 (O)  |
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